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Why are blunt noses better (subsonic)?
Compared to a sharper conical nose, a blunter conical nose is aerodynamically superior in subsonic flight. My question is why. I've read a lot about this and allegedly:

A blunter nose accepts a wider range of angles of attack without sideways airflow across it, which would cause additional drag that is avoided.

It would appear that airliners have blunter noses because they fly at a wider range of AOAs and gliders have sharper noses because they fly at a narrower range of AOAs. However...

At optimal AOA, the form and skin friction drag is also lower. I don't understand why!

According to this, a blunt nose is aerodynamically superior even when flying at the optimal AOA, but then why do gliders, which obviously value aerodynamics above all, always have relatively sharp noses? According to some descriptions, a hemispherical cap is the optimal shape for decreasing form drag and skin friction drag, but I haven’t found any description of why the first is true and the second is often motivated with hemispherical caps having the lowest surface area to volume ratio, but a conical cap of equal height has a lower surface area relative to the radius! In other words, if we were to design a new airplane, why wouldn't we sharpen the nose closer to a point instead of blunting it closer to a hemisphere, only considering aerodynamics?
Edit: yes, this is a unique question because I’m asking why and specifically why form and skin friction drag of a blunter nose is smaller than for a sharper nose, challenging intuition.
Edit 2: basically, why does No 2 in the following have less drag than No 5?

According to the image (I’ve yet to read the source, I’m on it!) there’s less pressure drag, but according to an answer by aeroalias to a related question, there’s less friction drag:
Why/when is the blunt nose better?
Which one is it? Both? Which one is most relevant? Or are both irrelevant in reality? Peter Kämpf’s answer makes no mention of either alternative, only arguing about the AOA and stagnation point.

Not sure if it's a dupe, but https://aviation.stackexchange.com/questions/24414/why-when-is-the-blunt-nose-better provides a pretty good explanation.

The first answer mentions my first point above and the second answer mentions the second point, but writes in text that the skin friction drag is lower while supporting it with an image that says the form drag is lower. The former is supported by the erroneously wetted area argument and the latter is stated as fact without reasoning (I’ve yet to read Hoerner). Why is the form drag and skin friction drag of a blunter (maybe even hemispherical) nose smaller than for a sharper nose (even at optimal AOA) and why do gliders ignore it?

Not a duplicate. This question states awareness of the difference between subsonic and supersonic. It is a well stated question and does not deserve to be downvoted.

I didn't downvote, to be clear. I'm just saying that the answer I linked provides a very good explanation that partially clears up some conceptual questions. Any extra answers on this page should at least add to what @PeterKampf explains in his answer.

About that image: a) Is the first coeficient really meant to be negative? ----- b) the graph shows no noses that are both sharp-tipped and smooth-walled, so they do not really support your question - #2 and #4 differ in tip and wall

Yeah, I read the source and the first coefficient is actually negative. Also, yeah, I wish the transition from the sharp cone to the cylinder was smoother here.

Nose shape on fuselages
Fuselage noses of airliners hold the antenna of the weather radar, and a blunter nose will put the least amount of material in the way of the electromagnetic waves. On a three-dimensional body the shape of the nose is of less importance and sometimes used to express a brand image. In subsonic flow, sharp changes in the contour must be avoided: Here, air would have to change direction abruptly which requires strong pressure gradients. Therefore, curvature along any flow path should change gradually. This favors an elliptic body shape. If you want to have no jump even in the second derivative, use a lemniscatic function.
A hemispherical nose would have constant curvature followed by no curvature along the cylindrical portion of the fuselage. The sudden change in curvature at the transition between the rounded nose and the cylinder would require a sudden jump in pressure which causes more drag than the gradual reduction of curvature (causing a gradual increase of pressure) of an elliptic nose.

Edit 2: basically, why does No 2 in the following have less drag than No 5?

Because the sharp corner at the base of the cone in No. 5 will cause a suction peak which in turn will cause a jump in boundary layer thickness. The hemispherical nose of No. 2 is better, but best is the elliptical nose in No. 1.
A blunter ellipse will have lower overall surface but a steeper change in pressure - here, as so often, the optimum is a compromise which needs to include the Mach and Reynolds numbers as well as structural considerations. Since the optimum is flat, there is considerable space for individual solutions.

What if No 5 had a smoother “ridge” (cone to cylinder transition) but was still pointy? Say, if No 1 ended in a point?

The pointy tip would not "fix" the stagnation point to that point - rather, it would still move with the angle of attack, but the part of the flow negotiating this tip would exhibit a strong suction peak and "smooth out" the contour by means of a local separation. This would place more strain on the boundary layer and increase its thickness downstream. Any hope for laminar flow over this nose past such a tip would be destroyed by this, of course.
The smoother ridge would be better, so the elliptical nose with a pointy tip would improve things like No. 4 does over No. 5. On No. 4 the sharp ridge will most likely produce a separation on the ridge which will make the whole body look larger to the oncoming flow. The bisection of the drag coefficient in No. 4 indicates to me that the separation is much smaller. Note that this is all only valid in subsonic flow! In supersonic flow No. 4 would look best.
Nose shape on wings
On wings, however, the nose shape is extremely important - witness the jumps im performance that sometimes were achieved by optimizing the nose shape. To get to the bottom of the wing nose shape, we need to talk about boundary layers, Mach effects and much more, so be prepared for a long answer.
Glider airfoils can afford to use small nose radii because they fly at low Mach numbers. Airliners, on the other hand, need to keep local Mach numbers low which favors a larger nose radius. If you are happy with this answer, better stop reading now.
At the Symposium Transsonicum of the International Union of Theoretical and Applied Mechanics (IUTAM) in 1964, E.V. Laitone presented something like a magic number for transsonic flow:
$$Ma = \sqrt{\frac{1}{\gamma - 1}} = 1.581$$
with $\gamma$ the ratio of specific heats of a fluid. Once the local suction peak at the nose of an airfoil reaches such a speed, lift stops growing any further. With stall speeds of around 120 knots, the early jet airliners routinely hit this limit using the 6-series NACA airfoils of that time.
Below you see the pressure distribution on a flapped airfoil near maximum lift (picture source). The specifics of the airfoil don't matter much; important is the suction peak at the nose which is made possible by a large angle of attack and a flap with a ventilated gap between wing and flap.

In such a case, a small nose radius creates a very strong but narrow suction peak because a very strong pressure gradient is needed to force the flow around the tight nose contour. If the nose is blunter, the suction peak can spread out lengthwise and becomes less peaky, simply because the trajectory change of the flow around the nose is becoming more gradual. Since there is a direct relationship between local suction and local Mach number, those suction peaks must be spread out and flatter in order to allow higher angles of attack. That is the main reason why supercritical airfoils have blunter noses - they will tolerate higher lift coefficients made possible with powerful flaps.
Initial glider airfoils used rather blunt noses and high camber. With composite technology the smoothness of wings could be improved and early numerical codes helped to shape the pressure distribution such that a fairly wide angle of attack range can be covered without incurring sharp suction peaks on either surface. The Eppler code of Richard Eppler was the first such tool and by prescribing pressure levels over sections of the airfoil at specified angles of attack it made laminar airfoils easy to design on the low-power computers of that time. The result are contours with a small nose radius which would produce a sharp suction peak once the specified angle of attack range was exceeded. By suppressing the suction peak over that specified angle of attack range, the laminar bucket could be maximized in a way that a blunter nose would not allow.
Now for that promised excursion into boundary layer theory: A laminar boundary layer is stabilized by a positive speed gradient (accelerating flow) but a negative speed gradient will trip the transition to turbulent flow rather quickly, especially at higher Reynolds numbers. The back side of a suction peak has just such a negative gradient; therefore suction peaks must be avoided in order to keep the flow laminar over much of the wing chord of gliders. In addition, using camber flaps allows to shift the laminar bucket up and down on the lift coefficient scale, so the combination of a small nose radius and a camber flap will allow laminar flow over a large lift coefficient range.
Airliners cannot do the same: The Reynolds number on their wings is so large that turbulent transition happens near the nose regardless of the local pressure distribution. Therefore, they are not concerned with laminar buckets and will happily tolerate the suction peaks that come with higher angles of attack. In cruise, when managing the pressure distribution is essential to reduce shocks, the angle of attack range is extremely narrow, so the blunt nose produces no disadvantage. However, this transsonic pressure distribution demands very little camber in the forward part of the airfoil, so in order to tolerate high angles of attack during approach and landing, they need a blunt nose. The larger nose radius also helps to integrate leading edge devices like slats and Krüger flaps, which is another advantage at low speed.

I’ve been praying for an answer from you :o) Written many good answers! I’ll dig into this...

Hold on, my question is about the nose of the fuselage and the answer appears to be about the nose of the airfoil, unless the answer is the same then? Basically, internet answers seem to suggest that blunter noses are aerodynamically superior to sharper noses for, say, an airliner based on 1) acceptance of greater range of AOAs, 2) smaller friction drag, 3) smaller pressure drag. I don’t understand why smoothing a shape decreases the friction and pressure drag, if that’s valid. And whether/why a hemispherical cap is aerodynamically ideal.

@Aerocurios: In this case maybe this answer is more fitting. Airliners need blunter noses to cover their weather radar. Aerodynamically they don't make much of a difference. Airfoil noses, on the other hand, deserve a longer answer. Sorry to have disappointed you.

As far as I’m aware, Airbus and Boeing noses are different because Boeing is an older design avoiding curved windows for cost efficiency while Airbus can create such windows with ease and as a consequence avoids the formation of a draggy supersonic bulb on top of the airplane. Newer Airbus/Boeing designs don’t have this difference anymore. But that’s extremely specific to Airbus and Boeing. I’m more interested in what would happen to friction and pressure drag in general if we made a nose (of say, only for instance, an Airbus) perfectly conical without changing its length!

In that case, would the friction and/or pressure drag increase or decrease and which one is greater in absolute terms? Or are both irrelevant and the only reason for blunter noses is that they accept a greater range of AOAs?

@Aerocurios: I immediately understood your question as one about wing noses because gliders have sharp and airliners have blunt noses. With fuselage noses there is no such clear distinction because nose shape is of minor importance. Both Airbus and Boeing use flat window panes on older designs; Boeing started to use curved windows on the 747 (more than 50 years ago).

Well, it’s important enough for no one to fly with cylindrical noses. If it really wasn’t important I would expect some airplane to have a trunk and tusks to amuse the passengers. Instead Airbus and Boeing airliners recently appear to have converged on a common almost exact contour. According to my interpretation, Boeing planes used to have a distinctly more two-stage nose (on top of it) and based on what I’ve read this was related to the cost of window curvature at more than one angle necessary to make them flusher, check 707 with 777 etc. However, maybe it was for other construction reasons.

Thanks a lot for continuing to update this! What if No 5 had a smoother “ridge” (cone to cylinder transition) but was still pointy? Say, if No 1 ended in a point?

OK: one final question: what about the skin friction drag? Which shapes would have the lowest friction drag? The short cone with the smaller wetted area or the elliptical cone, which I guess has a smaller boundary layer? I can’t guess which.

@Aerocurios: Skin friction drag is roughly proportional to surface area. The strain which a suction peak puts on the boundary layer actually lets friction decrease past the peak, but the high relative speed inside the suction peak more than makes up for this. So the smooth speed change of the ellipse also produces the lowest friction drag over the same area.

Blunt noses are best at subsonic speeds because they provide the best shape for the air to get out of the way. From this site:

The speed of "sound" is actually the speed of transmission of a small disturbance through a medium.

The picture shows a single point traveling at a constant speed V, emitting small disturbances :). The air in front of the travelling point is forewarned, and is pushed out of the way isentropically, without losses. It is this part of the subsonic airflow that is key to the optimal shape: a parabolical one. The point pushes air out of the way spherically, while travelling at a constant speed.
If our travelling disturbance is not an infinitesimally small point but an actual 3D body, its optimal shape is the same: parabolic, then rounded off at where the cylindrical body starts creating an elliptical shape. With this shape, the air in front of the body can get out of the way in the most orderly fashion.

The streamlines in the picture above are at equal distances. The air moves out of the way, creating a lower static pressure which sucks the nose into the airstream. With a hemispherical shape, the streamlines are closer together at some places, creating a pressure increase that negates the initial lower pressure.
Notice that in your figure 20 the first shape is elliptical and actually has a negative $C_D$ for the nose only: it sucks itself into the airflow. None of the other shapes do, not even the hemispherical shape 2. With subsonic incompressible flow, it's what happens in front of the nose that creates lower drag, not at or beyond the nose.
That's all valid when the shape travels at a certain speed, at zero Angle of Attack.

At any other speed, the optimal parabolic shape is different, still a parabola though for an infinitely wide body, or an ellipse for a body with finite dimensions.
At any other AoA, it is very difficult to create a 3D elliptical body shape, and it would be different for any AoA. But a spherical one comes close, as your Figure 20 shows - the first bit of a parabola is close to a sphere anyway. The larger the sphere radius, the closer it is to the optimum. And a sphere is a sphere at any angle.

Edit How can the shape suck itself into the air in front of it?
The pressure disturbances warn the air to get out of the way. Bernouilli is valid for low subsonic flow:
$$p_t = p_s + \frac{1}{2} \cdot \rho \cdot V^2$$
or: total pressure = static pressure plus dynamic pressure. Far in front of the moving body, $p_t$ is the total pressure is environment static pressure. As soon as the air starts to move, local dynamic pressure increases and local static pressure decreases.
Edit2
Yeah, elliptical for finite body dimensions.

Oh, that’s an interesting derivation of the shape! But I don’t really understand it, is there a source with more details because where in the first image do you get a parabola? And why does the parabola suck itself into the airflow (google search only gave computer-related answers)? Once I understand that, the AOA/pressure drag part of my question is probably answered to my satisfaction.

The parabola is at the connection of the circle outlines. The 3D shape with a nose shape that matches that of what happens with a point disturbance moves the air out of the way isentropically (without losses).

Sorry, my reference books are old uni books in a foreign language.

A smooth blunt nose would in theory cause a smoother airflow over the surface area of the nosecone. In a subsonic situation of course where a spear shaped nose would be effective at cutting through the air in a supersonic situation.


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When airplane pilots begin a 10 minute wait timer when lightning strikes, is there a geometric reason behind it?
Preface
This question is taken from a similar one inside the mathematics stack exchange.  I followed the suggestion given in the comment by Jyrki Lahtonen and posted said question here.
Background
Recently, on a flight back from Charlottesville, NC to Logan Airport, MA, the Pilot stated this to his passengers over the intercom, I paraphrase:
As you can see out the window, we just saw some lightning strike within 3 miles of our plane. Because of this, they closed the ramp gates. Every time lightning strikes within three miles of our plane, we have to reset the wait timer to ten minutes.
This made me wonder if there was a mathematical connection between the distance of a lightning strike and the amount of time needed to wait for take-off to ensure another strike will not occur on the ramp.
Work
Let the "area that causes a delay" when lightning strikes be represented by a circle of radius 3 miles. In addition, let us envision a scenario where lightning strikes on the circumference of the circle, i.e. 3 miles away from the plane, which is the farthest distance from the place that will trigger a 10 minute wait.
The intersection of these circles is a clean venn-diagram shape, and results in an area of (9π)/3 sq.mi. or 3π sq. mi., which is about 9.42 sq. mi, and the ceiling of which would be 10.
Question
Given that the pilot stated that a wait time of 10 min would be triggered if lightning struck within 3 miles of the plane, and the area in between a strike on the circumference of 3 miles and the plane results in a ceiling of 10, is there any mathematical basis for a lightning strike wait time of 10 if given a "danger zone" of 3 miles?

I fail to see the relevance of mathematics to this question. 3 miles is almost certainly an arbitrary distance employed (and probably not that accurately tracked to be honest) to ensure the safety of ground operators. Are you really asking why they close for another arbitrary amount of time?

The position of individual planes does not matter, and it is not the pilot deciding to wait. It is the airport that has rules regarding not having people work on the ramp and passengers walking around outside while a thunderstorm passes over the airport

No, because adding candles to candy does not make much sense.
The dimensions are different, so the relative numeric values depend on the choice of units. You could equally state the rule as you have to wait ⅙ of an hour after lighting strike within 5 km (if it's statue miles; 5.5 if nautical) and now the numbers won't match any more.
Besides, it is a 3 mile radius, so the area is about 28 square miles and the numbers never matched to begin with.

The real reason is two-fold:

When there is active storm right over the airfield, somebody could get struck by lightning. Inside aircraft and vehicles is safe, because the lightning will ground through the skin and not get inside, but out on the ramp it wouldn't be. 10 minutes is long enough to consider the storm no longer active.

The storm builds up static charge on the aircraft and vehicles. This is again not a problem for the occupants, because the conductive skin ensure the whole vehicle is at the same potential, but it could cause a static discharge strong enough to hurt someone or set something on fire when touching the vehicle from outside. 10 minutes is considered long enough for these charges to gradually discharge through the wheels to the ground and through the static wicks to the surrounding air.
This would also apply to aircraft that flew near the storm just before landing, but that is normally avoided as storms generate dangerously strong wings.

The 3 mile radius is simply size of typical storm.

There most certainly is mathematics behind this 3mile/10minute rule you encountered. This rule likely comes from statistics, and it's purpose is to give a simple rule ro follow, and thus avoid increasing the risk of getting struck by a lightning and/or a microburst.
I bet you can twist this statistical model into a geometric form, but i think it will be a 2d curve of normal distribution, not a 3d form.
P.S. I beg to differ with a commenter above who stated that *3 miles is almost certainly an arbitrary distance...". There's hardly anything arbitrary in aviation rules, regulations and procedures. Most things, if not all, have been learned the hard way...

It is arbitrary in the sense like 40 mph speed limit is arbitrary (rather than a mathematically optimal, say, 43.2). If the ligtning strikes at 3.1 mi, there is suddenly no danger? How did they choose the risk threshold? It is almost certainly a standard practical rule with 'typical' parameters which are easy to estimate and remember. Aviation has lots of such simplifications.

How is 43.2mph mathematically optimal? Aviation parmeters are simplifications, yes, but not arbitrary ones. In his answer Jan explains this case further.

If it's not arbitrary (as in "3 miles gives us roughly enough safety margin") then please demonstrate why it is 3 miles and not 2 or 4, or 10. Lots of things in aviation are arbitrary in that sense, in that they give enough safety margin without being able to calculate the exact figure required. eg, wind correction angles, descent profiles, course corrections etc etc etc.

We seem to have profound disagreement on the meaning of the word arbitrary. I understand it as: https://dictionary.cambridge.org/dictionary/english/arbitrary

@Jpe61, suppose that we somehow scientifically determined that the optimal speed limit in this place was 43.2 (it is unlikely to be a nice round number). But instead of posting this limit, we instead round it to an arbitrarily selected set of standard numbers (for various reasons, convenience basically). The set is arbitrary, because, in the first place, that the decimal system we use is arbitrary and is no more valid than an 11-base system, mathematically speaking.

Rounding 43.2 to 40 is not arbitrary.

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How do the engines of a multi-engine airplane shut down after fuel starvation?
If a multi-engine aircraft runs out of fuel, can the engines flame out at different times due to fuel inequality between the tanks, or do they both die at the same time because fuel is constantly balanced?

Thanks for the replies. In the case of the possible track of MH370 there is very good correlation between a constant track, constant speed and the Inmarsat ping ring data at specific times. That is until the distance between the 5th and 6th arc which, if the speed is kept constant, the distance is too short. I suspect that an engine shut down during this leg and the aircraft slowed whilst still being kept at height and on track by the auto pilot.

Be careful not to confuse "starvation" with "exhaustion" - Starvation is defined as "engine(s) not getting fuel" whereas exhaustion is "tanks have run dry".  Fuel starvation can happen with plenty of fuel remaining (John Denver's crash comes to mind).

On multi-engine aircraft the engine on one side is normally fed by the respective tank, so the left engine(s) is fed by the left tank etc... it is possible to cross-feed to allow access to all the fuel in the event of an engine failure.
In order for a simultaneous run down to occur - both tanks would have to have precisely the same amount of fuel loaded and both engines would have to burn fuel at exactly the same rate, which is unlikely.

While it is unlikely, starting with the same amount of fuel and consuming it at exactly the same rate is not the only way to get to both running out at the same time. There are, of course, an infinite number of possible ways to get to the point of them running out at the same time. It just there are a lot more ways that doesn't happen.

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Why are the transition altitudes/levels different around the World?
I got tasked by my instructor pilot to find out why each country has a different transition altitudes / levels. For example in the US it's FL180 and in Saudi Arabia it's FL150.
Can someone help me with that, and if possible provide any document that talks in details about this subject.

Just a note that your examples are transition levels, not altitudes. For your study, have a think about how things like terrain and pressure would affect the decision on which numbers to choose.

Related

Flight Levels are easier to work with for ATC and for pilots since everybody will use standard pressure setting. Effectively it means less work load and lower probability of making an error with a safety impact.
Typically the highest terrain elevation plus some margin is used as transition altitude. This gives the highest number of flight levels available.
For example in the Netherlands, being a flat country partly below sea level and the highest obstactle at about 1500 ft, has a transition altitude of 3000 ft for IFR flights and 3500 ft for VFR flights.
There is (was?) an initiative to create a common transition in Europe, but I haven't heard anything about it for a while. Some interesting reference documents were produced:
Towards a Common Transition Altitude - A Flight Deck Perspective
A Common European Transition Altitude  - An ATC Perspective
(both files in PDF format)

The highest peak in Saudi Arabia is Jabal Ferwa at 9,856 feet above sea level. Add some padding for safe terrain clearance, instrument error and human error, and 15,000 feet looks like a reasonable transition altitude.
The highest peak in the contiguous United States is Mt. Whitney at 14,505 feet, so 18,000 feet for the transition altitude looks similarly reasonable. (Special rules apply in Alaska around Denali at 20,310 feet.)
In theory, the transition could be much lower for non-mountainous areas of both countries, but apparently their authorities believed that a uniform transition altitude was simpler and therefore less prone to errors.

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Were any composites used on the A340 series?
Were any composites used in the construction of the A340? If so, which ones?

Yes!!
The slide shows that composite are used in rear pressure bulkhead and keel beam in addition to other parts.

Source: https://company.airbus.com/dam/assets/airbusgroup/int/en/investor-relations/documents/2008/presentations/GIF2008/gif2008_workshop_composites_hellard.pdf
You can read more:
http://www.lcc.mw.tum.de/fileadmin/w00bkg/www/PDF/Symposium/LCC_Symposium_Havar.pdf


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How many hydraulic systems does the Boeing 757 series have?
How many hydraulic systems does the B757 series have, does it have a Power Transfer Unit (PTU)?
What is the operating pressure of the hydraulics in the Boeing 757?
How big is the hydraulic reservoir?

The Boeing 757 has three fully independent hydraulic systems. The three systems are "Left," "Center," and "Right."

The left system is powered by an engine driven pump (EDP) on the left hand engine, an alternating circuit motor pump (ACMP) located in the wheel well, and a power transfer unit (PTU) powered by the right hydraulic system.
The center system is powered by two ACMPs located just aft of the wheel well in the wing to body fairings, the center system can also be powered by a ram air turbine in an emergency.
The right system is powered by an EDP on the right hand engine and an ACMP located in the wheel well.

All three systems operate at a nominal pressure of 3000 psi.
Each system has its own reservoir. The reservoirs are air-on-oil type reservoirs and use engine bleed air to pressurize the fluid to provide proper head pressure for the pumps. The left and right reservoirs have a full fluid capacity of 6.6 gallons (on the ground), while the center reservoir is 3.5 gallons (on the ground). The reservoir itself is larger than these full volumes (roughly 50% larger) to allow for various effects, like thermal expansion of the fluid and different exchange volumes in actuators.
In addition to the 2 EDPs, 4 ACMPs, 1 RAT, and 1 PTU, there are also two small electrical pumps in the left system that are used to unlock the uplocks on the landing gear for alternate extension. These pumps only function in flight when alternate extend is required, but they are also used to release the gear doors on the ground for maintenance using switches on the belly.
Any one hydraulic system can fully control the aircraft, and further any one pump on any one hydraulic system can fully control the aircraft.
The image below gives a good block diagram of the 757 hydraulic systems. Another good resource for learning about the configuration of hydraulic systems on commercial aircraft is SAE AIR5005 "Aerospace - Commercial Aircraft Hydraulic Systems."

Old question. Nice answer!

According to this, the 757 has 3 hydraulic systems and 2 PTUs

Or maybe the single PTU is composed of two sections, a motor on the right system, a pump on the left system. From Boeing: "The PTU is a hydraulic motor pump which transfers hydraulic power from the right system to the left system. The PTU is automatically activated if the left engine fails or the left engine–driven pump pressure is low.".

@mins I was thinking that. All Airbus aircraft to the extent of my knowledge have one PTU.


text:
Did Boeing subcontract the manufacturing of the wings on the B757 series?
If Boeing did subcontract, which company took over the manufacturing of the wings? Where were the wings made?

The 757 Wings and Wing & Center Body Section was manufactured at the Renton plant at KRNT.

Is the "Center Body" not the fuselage?

@AnandS oh, ok. Thanks for the clarification.

Winging center body is a section of the fuselage, which contains a structural members that the wings attached to. That is Boeing’s particular way to manufacture an airliner.


text:
Why stirring the cyclic helps in stuck left pedal?
In demonstrating stuck left pedal from a takeoff with my instructor yesterday when starting a nose left turn he had me stir or move the cyclic clockwise in small steady circles and pump the collective. It did slow the spin down significantly but I just don't understand why.
Any thoughts or comments would be greatly appreciated.

Which helicopter? Was there forward speed or were you in a hover?

The r22 and it was in a hover

The R22 has an anti-clockwise rotor. The more torque the rotor exerts, the higher the clockwise torque on the helicopter will be: nose right. With a stuck left pedal the nose wants to turn left, so we want the rotor to exert more torque.
On pumping the collective: there is an effect of how fast the collective pitch is raised, on the torque exerted. The figure below shows $C_T$ (a dimensionless measure of rotor torque) against time for a full sweep of the collective from full down to full up. Rapid deflection causes a torque overshoot, so a higher torque than a slow deflection.

So if the collective is raised fast and lowered more slowly, a net torque increase is generated while lowering the helicopter. Picture source: Helicopter Performance, Stability and Control by Raymond Prouty
On rotating the cyclic: cyclic pitch deflection causes angle-of-attack variation of the rotor blades, with an associated variation in blade drag, and therefore rotor torque. There is a section on this effect in the book Principles of Rotor Aerodynamics by J. Gordon Leishman (section 8.15.7 Unsteady Axial Force and Airfoil Drag). What I can distill from this chapter is that stirring the collective clockwise creates more airfoil drag, while keeping the net movements fwd/aft and left/right neutral. Unfortunately the section does not contain any helpful diagrams.

Thank you so much for the explanation! I have some studying to do on it in order for it to completely sink in but I can't say thank you enough Koyovis!

No worries Wes, wishing you lots of joy riding helicopters.


text:
How to detect landing and takeoff using the value of GND field in ADS-B messages?
How to understand that an aircraft is landing or taking off by analyzing MSG messages? According to this article we just need read GND field (22). I also heard that during landing the aircraft is sending MSG.2 with field 22 at 1.
I don't understand whether the aircraft is landing or taking off when the field is 1. It seems it just means the aircraft is on the ground, but there is no information about it's exact state.
How to actually detect that an aircraft is landing or taking off?

There is no field in any of the eight message types that indicates that an aircraft is in the process of either landing or taking off. Field 22, where it is transmitted, is a simple binary status: the aircraft is on the ground, or it is not. If you want to know what an aircraft is doing look at values for field 22 in successive messages. When the state changes you'll know that an aircraft has transitioned from ground to not-on-ground and has taken off (or vice versa)
Depending on how you define it, the actual period for which a 'taking off' status would be valid could be very short. I.e, just the duration of the take-off roll. Before that the aircraft is taxying and on-ground; after that it is flying. The same limitation is true for landing, but in reverse.

Thanks!
There are only two values 0 and 1, right?
Value 1 means, that  aircraft is on the ground.
And 0, that aircraft is not on the ground.
And there is no 'special type of message' corresponding the situation, when aircraft is on the grond (i.e MSG sends only field 22, so may be this type of message broadcast more frequently?)


text:
What is the most airliners shown flying together in a single photo?
I was re-visiting this question about a timelapse photo of multiple airliners and started wondering about the maximum number of jetliners that could be photographed together while flying.  I have found cases of Airbus aircraft flying in formation for promotional purposes, (see 5 here  and 6 here) and I'm sure Boeing have done the same.  However I was thinking that a busy airport with a lot of aircraft in holding patterns waiting to land might provide an opportunity for more, even though they would be maintaining vertical separation.  Does anyone know of photos like that?
To be a little more precise in what I am asking for, I am looking for a non-edited photo showing the most airliners where the aircraft must be originally built to carry passengers commercially and must be flying (not one of the many reasons photos of planes in "storage" on the ground or stacked up for demolition)

By passengers do you mean people paying an airline?  Otherwise any airplane with more than one seat would qualify, and events like EAA Airventure's anniversary reunions would be examples of many dozens of airplanes in quite close formation.

@CamilleGoudeseune I meant commercial airliners.  I've edited it slightly to make that clearer

Interesting question... I'd look for a photo of Denver or Atlanta or some other place that can run triple simultaneous parallel approaches.  A nighttime photo looking at the final would show landing lights out to maybe 20 or 30 miles (not talking time-lapse here, just that lights at night show up better than specks do in the daylight). Under the right conditions you might see 15-20 jets, though it'd take considerable patience to get the shot!

Someone posted an image here with 5 or 6 aircraft visible on approach. I'll see if I can find it.

You might look to Italo Balbo's Crociera aerea del Decennale mass formation of Savoia-Marchetti S.55s, which in 1933 consisted of 25 aircraft touring North America.
There are a large number of photos and newsreel of the occasion - the most I can find in a single picture is twenty-one aircraft flying over the Statue of Liberty.

source
This fits your criteria, as the S.55 was an early airliner "originally built to carry passengers commercially" - each hull could accommodate four or five passengers, with hammocks in the rear. However, these aircraft were being flown to demonstrate Italy's aviation prowess by military pilots, and the S.55 was also used by Italy as a warplane.
For an entirely commercial formation of the same aircraft type I'd suggest the Concorde formation of four aircraft that flew on Christmas Eve 1985

source

Exquisite discoveries!

This old question got bumped recently, but piqued my interest.
There's a fair few stock photos of stacked airliners, but no source quoted so they could be composites.
I did however come across a YouTube of someone filming approaches into Heathrow on one particularly busy period.  There's frequently 4 airliners in shot, obviously more would be visible had the field of view been wider.

The image found here shows 13 DC-3's in formation which were used as airliners at one point but these are in military livery in the photo.
Here is the kind of photo you describe but only 3 aircraft are in view. Keep in mind aircraft are generally turned onto the final approach path and only a few may be lined up straight off the runway at any given time.
The airbus operation you link, may be the biggest instance of passenger jets flying in tight formation.

The DC-3 pictures are really cool, particularly the one showing them from the ground.

Yeah, thanks for sharing that D-Day anniversary link.


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How will the Boom SST not require an afterburner?
The company Boom is developing a supersonic airliner (SST).

In one of their blog-entries1 they state they would not need to use after-burners because they use modern turbo-fan engines.

Does this mean they will use a variable geometry exhaust?
Is there another (efficient) way to become supersonic with turbo-fan engine?
Which turbo-fan engine are they planning to use?

1: blog.boomsupersonic.com/why-we-dont-need-an-afterburner-a4e05943b101

well for what is's worth @mins, they have Japan Airlines and Virgin on-board.

MAYBE they’ll pull it off; most likely the program and company will go bankrupt on this venture.  Always beware of Silicon Valley whiz kids touting this kind of stuff.

All I have to do is look at the specifications on that aircraft and realize that if they could bring it into fruition,  it will be a good engineering achievement but a commercial lemon. They have a limited range of 4500 nautical miles - I don’t know if they can maintain that during supersonic cruise - so scratch trans pacific routes where the big bucks would be for an SST.   Regulations in Europe and the Americas prohibits supersonic flight over land, so that eliminates even more routes.   It also has a balanced field length listed at 8200 feet.  That’s s hell of a long takeoff roll.

Is this 4500 limit due to refueling? What does tech stop mean?

The business of those companies are basically: 1) make a good presentation, get funding, 2) spend the funding, convert cash to the "value of the company", 3) sell the company, convert "value of company" back to cash. "entrepreneur"s make money at step 3. Individuals, consultants, contractors make money at step 2. There's no reason to force yourself believe their design makes sense because it doesn't have to make sense in the first place.

wow, I am a bit surprised that I triggered so many negative/cynic comments.  Especially since I would argue that my questions are not even related to the  commercial facts.

What makes you think afterburners are necessary for supersonic cruising? Several modern military fighter jets achieve this (F-22, to name one), they call it "supercruise", and specifically designed the airframe to not require afterburners for Mach 1+ supercruising. These jets usually use turbojets, instead of turbofans, although I think the F-22 and F-35 use turbofans.

@SnakeDoc, these days, fighter jets all use turbofans, though the bypass ratio is low. But I don't think any of them is a pure turbojet these days.

@SnakeDoc the F-22 and F-35 use convergent-divergent nozzles together in their very low-bypass-ratio turbofans.

Concorde too had a few airlines onboard (including JAL)

I do not think that not needing an afterburner has as much to do with their engine choice as it does their aerodynamics. They think that between (1) the area ruling they will do with the fuselage, (2) the chine that will allow them to achieve stability with an overall smaller wing, and (3) the wing itself with more sweep on both the leading and trailing edge which reduces induced drag, should altogether substantially reduce the amount of energy required to pass through the transonic regime, which hopefully will mean that they shouldn't need to use an afterburner.
I wish them good luck in this whole endeavor.


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Does a deflecting control surface change the wing area?
It might seem like a silly question, but bear with me.
I am using the standard formula to calculate the lift force of a wing.
L = (1/2) d v2 s CL
My question related to the s part of the formula. According to what I can find this area is calculated as if the wing was completely flat. This would be from the leading to trailing edges.
Wing Area = Wing Span x Average Chord

I am currently doing exactly this, to calculate the lift force of the wing on a computer model while model is flying. It is done at regular intervals along the wing span and added up. The sections that include control surfaces have different chord lines when the control surfaces rotate as expected causing the desired affect on lift.
BUT I noticed that the wing area also changes with the change of control surface angle. As the control surface moves closer to its upper or lower limits the surface area changes too.
Control surface at 0 deflection and at max
Red part is a control surface.

Obviously on a simple wing, the deflecting control surface cannot affect the (3d) surface area of a wing. Or should not. Yet using the formula to calculate the s part the wing now has a smaller area.
To get to the point, I would like to know if this makes sense. Does the ...effective wing area change during operation of a control surface ?

This is like "does speed change the meaning of time" or "does inflatation of a tire to different pressures affects its weight". Answer is "yes but depending on what you are working on it may or may not be meaningful"

TL; DR: It depends on the specific assumptions of the aerodynamic model, but normally wing area is considered constant.
Wing area normally is defined as fixed, and the CL is given so that any control surface deflection results in a change in lift curve, i.e. CL, only. One could equally attribute the change in total lift following a control surface deflection to an associated change in surface area, and then the resultant new CL to go with the deflection would accordingly be of different magnitude, so as to yield the same total lift.
It is a convenient general convention as far as I know, to consider wing area (also very appropriately dubbed as reference area) as constant, because it essentially eliminates one variable from the model.

I’d add that it’s called “reference area”, which also simplifies comparing different shapes for their efficiency.

@GürkanÇetin Good suggestion - done!

Capt @CptReynolds, this thing is quite confusing me. In Wikipedia. It is written like this: "The aspect ratio AR is the ratio of the square of the wingspan b to the projected wing area, which is equal to the ratio of the wingspan b to the standard mean chord (SMC)". As surface area affecting the wing area, means that AoA is calculated as the factor affecting the lift force.

In my opinion, increase the AoA (before it get stall) definitely will increase thrust, which will accelerate the airplane, which at the end will increase the lifting force. But at the same time the increase of the AoA will decrease the lifting force. Mean, the increase of the AoA is not linearly increase the lifting force.


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In a jet, are Vy and Vx achieved at a constant AOA at all altitudes?
In terms of IAS, Vx (best climb angle speed) increases with altitude, and Vy (best climb rate speed) decreases with altitude, until they meet at the plane's absolute ceiling. http://cospilot.com/documents/Why%20Vx%20and%20Vy%20Change%20with%20Altitude.pdf
In a jet powered aircraft, is this achieved at a constant angle of attack (AOA) regardless of altitude and other conditions (temperature, pressure, etc.)?

Question was edited to be more clear, narrow and to-the-point. If you see fit, please vote to reopen.

The link works.  I get a 3-page pdf by Rod Machado, including four complicated plots.

I had better luck with this.  Worth reading.

It might help to derive both speeds from first principles. We assume that the drag polar of the aircraft can be described by a parabola, like this:
$$c_D = c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$$
The symbols are:
$\kern{5mm} c_D \:\:\:$  drag coefficient
$\kern{5mm} c_{D0} \:$  zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$  lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$  3.14159$\dots$
$\kern{5mm} AR \:\:$  aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$  the wing's Oswald factor
Next, we describe thrust $T$ over speed $v$ with $$T = T_0·v^{n_v}$$
Now first to the maximum climb angle. This is reached when the condition $$\frac{\delta \gamma}{\delta c_L} = 0$$
holds true. No change in lift coefficient $c_L$ will improve the climb angle $\gamma$, it is only downhill from here to both sides. In order to get a grip on the climb angle, we look at the force equilibrium in steady flight at full power, assuming small values for $\gamma$:
$$sin\gamma = \gamma = \frac{v_z}{v} = \frac{T - c_D\cdot \frac{\rho}{2}\cdot v^2\cdot S_{ref}}{m\cdot g} = \frac{T_0·\left(\sqrt{\frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot S_{ref}}}\right)^{n_v}}{m\cdot g} - \frac{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}}{c_L}$$
The symbols are:
$\kern{5mm} m\:\:\:\:$ aircraft mass
$\kern{5mm} g\:\:\:\:\;$ gravitational acceleration
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} v_z\:\:\;$ climb speed
$\kern{5mm} S_{ref} \:$  wing area
Ideally, we would also multiply the climb angle with an acceleration factor, but I leave this out here for simplicity.
Now we can derive the expression for the climb angle with respect to the lift coefficient and get
$$\frac{\delta \gamma}{\delta c_L} = -\frac{n_v}{2}·c_L^{-\frac{n_v}{2}-1}·\frac{T_0·(m·g)^{\frac{n_v}{2}-1}}{\left(\frac{\rho}{2}·S_{ref}\right)^{\frac{n_v}{2}}}+\frac{c_{D0}}{c_L^2}-\frac{1}{\pi·AR·\epsilon}$$
The general solution is $$c_{L_{{\gamma_{max}}}} = -\frac{n_v}{4}·\frac{T·\pi·AR·\epsilon}{m·g}+\sqrt{\frac{n_v^2}{16}·\left(\frac{T·\pi·AR·\epsilon}{m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$
For jets ($n_v = 0$) the solution is quite simple, because the thrust terms are proportional to the thrust coefficient $n_v$ and disappear: $$c_{L_{{\gamma_{max}}}} = \sqrt{c_{D0}·\pi·AR·\epsilon}$$
For turbofan and propeller aircraft, we have less luck and get a much longer formula. This is the one for propellers ($n_v = -1$):
$$c_{L_{{\,\gamma_{\,max}}}} = \frac{T·\pi·AR·\epsilon}{4·m·g}+\sqrt{\left(\frac{T·\pi·AR·\epsilon}{4·m·g}\right)^2+c_{D0}·\pi·AR·\epsilon}$$
So yes, pure turbojets have an optimum lift coefficient for maximum climb angle that uses only terms which are constant over altitude. They do indeed climb steepest at a constant lift coefficient.
But the thrust-dependent optimum for other engine types hints at an altitude dependency which might affect the other optimum, that for best climb speed.
In order to find the conditions for maximum climb speed, repeat the process above with an expression where both sides are multiplied by speed:
$$v_z = \frac{T\cdot v - c_D\cdot \frac{\rho}{2}\cdot v^3\cdot S_{ref}}{m\cdot g} = \frac{T_0·\left(m\cdot g\right)^{\frac{n_v-1}{2}}}{\left(c_L\cdot\frac{\rho}{2}\cdot v^2\cdot S_{ref}\right)^{\frac{n_v+1}{2}}} - \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot c_L\cdot S_{ref}}}\cdot\frac{c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}}{c_L}$$
Now the solution for turbojets becomes the more complicated one, but that needs to be the case - how else would those optima converge at altitude?
$$c_{L_{{\,n_{z_{\,max}}}}} = \sqrt{\left(\frac{T·\pi·AR·\epsilon}{2·m·g}\right)^2 + 3\cdot c_{D0}·\pi·AR·\epsilon} - \frac{T·\pi·AR·\epsilon}{2·m·g}$$
While the angle of attack for steepest climb is constant over altitude, the angle of attack for best climb speed increases as excess thrust disappears with rising altitude. Thus, the question can be answered: No.

...constant AoA at all altitudes

Well, let's throw in a third propulsion system, a rocket, and compare thrust output vs altitude.
Excluding Mach effects, constant thrust means constant IAS means constant AoA.
Vx and Vy do not converge until decreasing thrust forces the aircraft to seek its lowest drag airspeed, between Vx and Vy.
We can see by drawing thrust vs airspeed curves for all three what happens as altitude increases.
The piston/prop has two strikes against it as propeller RPM must increase and the turbocharger has to work harder to maintain mixture ratio for more fuel.  At some point, available thrust decreases, then Vx and Vy must converge at the absolute ceiling.
The turbojet, operating on excess air, can dump in more fuel to power its compressor and maintain thrust a little longer, but it too reaches a point where thrust begins to drop off, and Vx and Vy converge.
The rocket never experiences a loss of thrust, therefor its Vx and Vy AoA remain constant.
The key is: constant thrust at constant IAS gives constant AoA.  Both Vx and Vy are higher drag configurations than Vbg, one for higher climb angle, the other for higher climb speed.  They can only be flown if there is adequate thrust at their respective IAS.

Speaking of thrust, the aircraft could hold Vx or Vy IAS and AoA as thrust reduces by reducing their pitch to the horizon, thereby reducing the gravity component acting against thrust.  In all cases the plane has 0 excess thrust available at absolute ceiling. It can no longer climb.  Flying this profile, the IAS would not converge until the very end.


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How can an up-lifting tail have normal force direction and gradient when generating UP force?
This article includes this statement about tail forces to validate the concept that the horizontal tail in conventionally tailed aircraft can be generating force DOWN or UP at different times and still have normal stability and control characteristics:

I took a Cessna 172 Skyhawk and put a couple of large pilots in the
  front seats, with no luggage and no other passengers. That meant the
  center of mass was right at the front of the envelope, so the tail had
  to produce considerable negative lift in order to maintain
  equilibrium. There was lots and lots of angle of attack stability. I
  took the same Skyhawk and put a small pilot in the front seat, a
  moderately large mad scientist in the back seat, and 120 pounds of
  luggage in the rear cargo area. That put the center of mass right at
  the rear of the envelope, so the tail had to produce considerable
  positive lift in order to maintain equilibrium. The airplane still had
  plenty of stability. (As far as the pilot could tell, it was just as
  stable as it ever was.) The easiest way to determine whether the tail
  lift is positive or negative is to observe the direction of motion of
  the tip vortices, as discussed in section 3.14. To observe the
  vortices, I attached a streamer of yarn, about half a yard long, to
  each tip of the horizontal tail, at the trailing edge. The streamer
  gets caught in the vortex, so its unattached end flops around in a
  circle. When the tail is producing positive lift, the circular motion
  is in the direction shown by the green “circulation” arrows in figure
  3.29, i.e. downward on the inboard side. When the tail is producing negative lift, the direction of motion is the other way, i.e. upward
  on the inboard side.

Numerous posts on ASE, and too many comments to count, support this concept.  That's great.  There is a nagging problem though that prevents me from accepting the "consensus" just yet.

For the tail to generate downforce, local flow must be above the zero lift line of the tail airfoil.  The hinge moment imparted by air loads on the elevator is TE down, control stick forward, and aft stick force is required to counter it.  For trim, the tab must move TE down to apply a servo force UP to counteract air loads.
For the tail to generate upforce (lifting), the local flow must be below the zero lift line of the tail airfoil.  The hinge moment imparted by airloads on the elevator is now TE UP, control stick aft, and FORWARD stick force is required to counter it.  For trim, the tab must move TE UP to apply a local servo force DOWN to counter act air loads.

This is what you would call "stick force reversal" and I can't find any resources that say this is a good thing; actually quite the opposite, and I certainly wouldn't want to experience it in my own airplane.
So how exactly does that work?  How can a lifting tail still have normal control force direction and gradients, same as a downforce tail, and how could the trim tab work given that it now has to work in the opposite direction to provide the required servo forces for trim, on the same aircraft?

Let me break it down in sections. It's rather long I'm afraid.
1. Tail lift direction
Consider the following diagram of a wing-body and tail combination, with a non-pitching aircraft in steady-state. Let's assume that a true wingbody mean aerodynamic centre ($h_{n_{wb}}\overline{c}$) exists and therefore the pitching moment does not vary with AOA at that point.

Image ref: Etkins, Dynamics of Flight
The pitching moment for the whole aircraft about every point must be zero. Let's pick the point of CG ($h\overline{c}$) to sum our moments, since it also corresponds to the point of rotation for a free-body (this equation can be found in Etkins, Dynamics of Flight, but its derivation is simple and you should be able to deduce it from first principle):
$$C_m = C_{m_{ac_{wb}}} + C_L(h-h_{n_{wb}}) - \frac{\overline{l}_t S_t}{\overline{c} S} C_{L_t} = 0$$
where $C_m$ is the total pitching moment, $C_{m_{ac_{wb}}}$ is the wingbody pitching moment coefficient about the MAC, $C_L$ is the total lift coefficient (wingbody + tail), $C_{L_t}$ is the tail lift coefficient normalized over the tail surface area ($S_t$), $\overline{l}_t$ is the distance from the tail MAC to the wing MAC, and $\overline{c}$ is the wing reference chord.
The sign of the tail lift, $C_{L_t}$, to make the above hold can be either positive or negative, depending on the relative contribution of each of preceding terms. Note two facts:

For a positively cambered wing, $C_{m_{ac_{wb}}}$ is typically negative.
$C_L$ is positive for 1G flight.

Consider some extremes:

If the CG is way ahead of the wing MAC ($h \ll h_{n_{WB}}$), then the second term will be very negative, and $C_{L_t}$ must be negative to compensate. That is, negative tail lift.

If the CG is behind the wing MAC ($h > h_{n_{WB}}$), and $C_L$ is very large, then $C_{L_t}$ must be positive to compensate. That is, positive tail lift.
Furthermore, if the CG is behind the wing MAC, then there will always be an AOA above which the trimmed tail lift becomes positive, assuming stall does not occur.

2. Neutral point
Neutral point ($h_n\overline{c}$) is the longitudinal position at which the aircraft does not experience any pitching moment with a variation in AOA or $C_L$ (in another word, NP = MAC of the aircraft). If the CG is ahead of the NP, the aircraft will experience a pitch down moment with increasing lift (statically stable); and vice versa for unstable.
I will cite the following result, once again from Etkins (but simplified):
$$h_n = h_{n_{wb}} + \frac{a_1}{a} \frac{\overline{l}_t S_t}{\overline{c} S} \left ( 1 - \frac{\partial \epsilon}{\partial \alpha} \right )$$
where $a=\frac{\partial C_{L}}{\partial \alpha}$ is the total lift slope of the aircraft, $a_1=\frac{\partial C_{L_t}}{\partial \alpha_t}$ is the isolated lift slope of the tail, $\epsilon$ is downwash from the wing on the tail.
Note that the second term is always positive. Therefore, the CG can be behind the wing MAC, yet ahead of the NP.
3. Elevator hinge moment and stick force
For a reversible aircraft where the column is directly connected with an elevator, stick force gradient is directly conferred by the aerodynamic hinge moment on the elevator (let's assume there is no down-spring or bob-weight for simplicity).
For a non-cambered tail, there are two main contributions:

Hinge moment due to flow incidence ($b_1$)
Hinge moment due to elevator deflection ($b_2$)
Hinge moment due to trim tab ($b_3$)

You are right to point out that if a tail is lifting, and its flow incidence increases with aircraft AOA, then there will be increasingly TE up HM on the elevator, which would imply instability. However, static stability ensures that there is increasing TE up elevator needed to trim for increasing AOA, which will add TE down HM on the elevator.
The details are more nuanced than the above, and it will involve airspeed since HM is squared to airspeed, but it can be shown that the stick force reversal only happens when the CG is aft of the stick-free neutral point ($h_n^{'} \overline{c}$), and not at neutral point:
$$h_n^{'} = h_n - \frac{b_1}{b_2}K$$
where $K$ is a positive constant that is a function of the aircraft geometry (I've simplified from eqn 2.6.13 from Etkins). As you can see, if $b_2$ (HM due to elevator deflection) overpowers $b_1$ (HM due to flow incidence on tail), then we pretty much recover the NP. Down-spring and anti-servo tabs are some methods to increase the $b_2$ artificially.
More specifically, please consult the following figure, adapted from Bromfield, Factors affecting the apparent longitudinal stick-free static stability of a typical high-wing light aeroplane and Etkins, Dynamics of Flight:

In the graph, $\delta_t$ is the trim tab deflection.
Observations:

The first two curves are with the tab always set to 0 incidence against the elevator. The movement of the CG purely shifts the stick force vs. V curve up and down. When the CG is aft of the stick-free neutral point ($h>h_n^{'}$), any steady-state speed would require a pull force (despite the stick may be forward of neutral if the CG is still ahead of the neutral point!)

The last two curves are with the tab set for a trim speed ($V_{trim}$). Now, when the CG is aft of the stick-free neutral point, the stick force gradient also goes unstable! What effectively happens when the CG coincides with the stick-free neutral point is that the hinge moment contribution from the elevator+tab cancels that from the tail AOA; therefore, zero steady-state force is needed to hold any particular elevator/stick position.

Either way, whether the tail is lifting doesn't dictate the stability of the stick force vs. airspeed.

Thanks for the trouble.My question remains however; if the tail is lifting, the local AOA is below the zero lift line, and hinge moment induced in the elevator by the local flow will be up, not down. I can't see how that doesn't result in a stick force reversal. You said the CG is ahead of the NP which makes a ND pitching moment, that has to be trimmed with downforce. If the tail is lifting up,it means that there is a NU pitching mmt that has to be trimmed with upforce.To have a NU pitching moment, the CG m b aft of the NP. Do all of you guys see the problem with trying to square this circle?

@JohnK Please see modified third section for your query.

Thanks again. Still filling some gaps in perceptions here. Can you describe the force changes that occur when the CG moves aft of the SFNP that leads from stability to instability?  If the tail can be generating lift upward with a statically stable condition, and the CG is shifted aft beyond the SFNP, what changed, beyond the tail having to lift harder than it was when CG was more forward. I have another question about NP, and why, if it is the average of the sum of all lifting forces, the tail area is included when most of the time its lift is down.  Perhaps I shld post a new question.

BTW the notification tag created with an @ ahead of your name doesn't seem to be working when I post comments.  No idea why.

@JohnK Re your first part: when the CG is at SFNP, then the relative HM from flow incidence and elevator are equal, and you need zero stick force to hold any particular elevator position at steady state speed and when trimmed at the beginning with a trim tab (again, no downspring or bobweight). I don't understand your second part.

and so that transition from positive to neutral to negative HMs,leading to positive to neutral to negative stability, occurs while the tail is generating upward lift,after the transition from downforce to no force to upforce (more or less the transition local AOA from above to below the tail chord line)?You see, in my mind, that transition coincides with the transition from downforce to, to zero lift, to upforce. That's the basic element I can't wrap my brain around.Books I have, and literature I've been able find online do a very poor job of explaining this beyond presenting formulae.

@JohnK That's correct. The forces themselves don't tell the story of stick force per V for reversible aircraft.

Ok a light bulb finally went off, I think. Could it be said that NP/CG static margin relationship refers to what I might call (an oversimplification but good enough for the concept) a "vertical weathervaning tendency"?  It has no direct bearing on the actual direction of forces, only how the variations in those forces creates restorative moments about the CG insofar as the forces are acting act an aerodynamic center that is some distance from CG, like a weathervane or vertical fin?  And therefore the upforce/downforce tail balance of forces is more related to dynamic stability?

@JohnK Nothing here has anything to do with dynamic stability. I've revised the post one more time. Hopefully this clears things up.

Ok great thanks!

Your thinking assumes that a specific elevator position produces a specific stick force. If I understand you correctly, a TE up elevator must be unstable because the  required stick force to bring it back to neutral is the opposite of what a TE down elevator requires.
That is not how it works. Neutral elevator is just one of many possible positions for trimmed flight*. Next, you need to think in stick position changes, not absolute stick positions. Mathematically speaking you need to look at the derivatives.
A stable stick force is one which brings the airplane back to its equilibrium after an upset. This might be a gust or a jolt on the stick; details don't matter. All that counts is if that airplane finds itself in a state different that what had been trimmed, the stick force must act in the direction which brings the plane back to the trimmed state.
Let's look at the TE up case first. The airplane suffers an upset which raises speed by something above trimmed speed. Wing AoA is reduced and so is tail AoA, so the downforce on the tail is increased over the already existing downforce. The aircraft raises its nose and slows down. Mission accomplished. (Yes, quietflyer and Guy, now a phygoid motions starts, but we are only looking at static stability here).
What happens to the stick force? The lowered AoA would drive the elevator from a TE up position towards TE down and the stick more backward, as it should for positive stability, until the old, higher AoA is restored.
Now the same with the TE down position: The change in AoA is the same, and so the elevator would be driven even more TE down. Again, the stick moves backward and shows stable behavior. There is no fundamental difference to the TE up case.
It does not matter where the elevator TE stands for trimmed flight. All that counts is the change as a consequence of a deviation from the trimmed state. The only difference between down- and upload on the stabilizer at a given speed is the amount of change. With a downforce the center of gravity is located more forward and the aircraft is more stable, so the change in stick force as a consequence of an upset is more pronounced. But the direction of the change does not change until the lift per area on the stabilizer is higher than that on the wing. But then the center of gravity will be behind the neutral point and the aircraft will become unstable.

* A naturally stable airplane with a fixed stabilizer will see the TE of the elevator rise as it is trimmed for slower speeds. At the same time, tail downforce decreases and becomes tail lift at low speed even as the elevator trim deflection is negative (TE up). The elevator (and with it, tail camber) runs against the required tail lift. This means tail lift is most likely to occur with negative elevator deflections.
A trim tab which produces mostly up forces could be combined with a spring in the elevator linkage in order to increase speed stability (read here how this works) or it serves to trim the airplane with flaps down because this case needs more negative elevator deflection. More down travel of the trim tab (meaning more trim range with elevator TE up) is completely normal and does not indicate that the tail will only produce a downforce over the speed envelope.
Stick forces are the difference in hinge moment between the free-floating elevator angle and the angle needed to trim the aircraft at that speed, multiplied by the elevator linkage gearing ratio. If the aircraft flies faster than what it has been trimmed for and the free floating angle is more positive (= TE down) than the trimmed elevator angle, stick forces will be in "pull" direction (the pilot has to actively push the stick away from the free-floating position in order to fly faster). This happens regardless of tail lift as long as the center of gravity is ahead of the airplane's stick-free neutral point.

Kampf  Thanks for the answer.  My point is actually about the elevator hinge moments present from airloads, and the 0-moment or "trail" angle the surface wants to seek absent a trim tab (in which case trim is solely a function of stabilizer incidence and if it's fixed, you have only one trim speed). If the tail lift is up, it means the local AOA relative to stab chord line is from below and there is a TE up hinge moment for most of the elevator's travel. If tail lift is down, local AOA is from above and there is a TE down hinge moment for most of the travel. Continued..

The way trim tabs are geared seems to accommodate the downforce case only, because most of the tab's travel is below elevator surface to provide an UP servo force, the more so that slower you are. So I'm not sure how a tab geared to be deflected mostly down for low speed can accommodate transitions from down force to up force at some point in the speed range.  I'm not saying you can't make a lifting tail work, I'm just not convinced the same airplane transitions between up and down lift somewhere in its speed range. If you want a lifting tail, you are creating a tandem wing airplane in effect.

@JohnK. Trim tabs are only one of several ways for trimming. Many small airplanes use a spring in the linkage which pulls the elevator up or down. Also please note the additional edit of the answer.

Does this issue exist on T tail gliders?

@RobertDiGiovanni: Yes, of course it does.

If I may ask again, the issue that John K was describing, down "float" on an uplifting tail TE (elevator down).  I had thought it may have something to do with the down wash of the 172 high wing.  A T tail flown in an aft CG (lifting tail) configuration may help solve this.

@RobertDiGiovanni: You are right that moving the tailplane up relative to the wing will diminish the downwash, but only by the cosine of the angle under which the tail now sits relative to the wing. Normally, a fuselage is much longer than the vertical is high, so the effect is small.

@PeterKämpf Thanks to both you and JZYL. I've gotten some great insights that one can't get from the usual resources.

@JohnK You are very welcome!

The last part is not entirely accurate. Please see section 3 of my post.

@JZYL Better now?

@PeterKämpf Thanks

Keep the wind flow the same.  Draw the chord line from each trailing edge deflection.
Indeed, this is how we switch from upforce to down force when we use our elevator.
This one twisted my brain a little too.  But we must consider the lifting torques around the center of gravity. If the AOA of the tail is lower than the wing (up or down), then the percentage change in lift created will be greater for an increase of AOA for the tail than the wing.
But even if all lift is borne by the main wing, an increase in lift (from increasing speed) will create a vertical departure from the line of flight, which creates downforce on the tail.
A sink from decreasing speed has the opposite effect.
So one can design the CG anywhere they want.

... on the same aircraft?

The problem comes from abusing the CG range.  With aft CG, the torque forces move towards instability, and the elevator and trim authority to counter act it are used to a greater degree as lifters rather than pitchers or trimmers.  Additionally the AOA ratio of wing to tail falls, further reducing stabilizing advantage.
The flip side of this is that too far forward a CG is harder to pull out of a dive, as now more pitch up authority is depleted.  A wing that has a center of lift that shifts with AOA adds to the design considerations.
So one can have a "staticly stable" lifting tail. The old fashioned draggy way is a low aspect delta with a lot of area, set at a lower AOA than the wing.  It's torque will be more stabilizing than the wing torque is destabilizing when pitch changes.

You didn't address the stick free force reversal of the elevator surface itself. I'm talking about a static condition "stick free", where the elevator is held to an angle by the trim tab.  I can push on the elevator and force the nose over, making the tail lift briefly, but stick free the elevator always wants to float DOWN, and either me applying back pressure, or the tab, applies the opposing force.  If it was lifting, stick free, the forces would be reversed.  This is what happens when the CG is way aft of the aft limit, behind the neutral point.

@John K this may be more of a localized phenomena at the TE than a change in relative wind.  I do know the 172 has some crazy things going on from the high wing downwash.  Normally, one would not expect the elevator to drift "down" if it were lifting.  I would try this (with all caveats for rear CG) at different airspeed (there for different AOA), making sure the plane was not rising or sinking, to further study it.  What was your airspeed when you did this?

No. Both the elevator and ailerons have hinge moments induced by the airflow. They naturally want to float in the direction of the moment. The ailerons see a moment TE up because the wing is operating at a posittive AOA and want to float UP.They are prevented from doing so by the linkage to the opposite aileron. Elevator is configured, on every tailed airplane I've even seen or studied or flown, with the trim tab configured to resist elevator DOWN float, not up float.If the tail was "lifting for trim", the elevator would be floating UP. This NEVER happens on any airplane with the CG in limits.

@John K Might be one for the wind tunnels.  Any direction of "float" implies more force on one side than the other.  If the tail were lifting, there should be more force on the bottom.  No one ever said the 172 is perfect.  It is good, but perhaps a bit close coupled.

No.  All conventionally tailed airplanes are the same. The force you feel on the stick is the induced moment applied to the surface by the airflow, wanting to make it float DOWN.  The trim tab applies the force for you from the tab trailing edge so you can let go and the slower you want to go, the more the tab has to displace.  Work the trim wheel on ANY airplane and slow=tab TE down.  If the tail was lifting, the tab on any airplane would have to work in the opposite direction to have the same function.  Anyway, thanks for the input but the response doesn't address the question.

@John K still working on it, and Bernoulli may have the answer.  The trim tab controls the elevator position  by aerodynamic force.  when the elevator is down, the trim tab is up, correct?.  Guess where the trim tab is?  In the accelerated airflow of the uplifting tail.  Perhaps not designed for this, now too big for the elevator, it pushes it down.  Might not even blame the main wing for this.  Notice, when the elevator is trimmed up, the trim tab is down, in the slower airflow.  In the wind tunnel, it could be checked.

The tab goes wherever you set it.  Set it to fly at minimum speed stick free, and the tab is full TE down, to drive the elevator surface UP against air loads.  At max speed the tab is TE up to let the surface trail and pretty much all the downforce is being generated by the fixed stabilizer.  If the tail was lifting, the air loads would be driving the surface the opposite way and the tab would have to reverse direction to do the same thing.  This simply doesn't happen. Fly a C-180 with a trimming stab, and the elevator trails all the time and all of the trimming force is coming from the stab.

And the stab is moving, which way when you spin the trim wheel? LE down to go slower, AT ALL TIMES. Because the local flow is always from ABOVE.

Well, I only have flown a 172.  But I do agree with "local airflow". It needs more study.

Just keep an open mind, and be prepared to dump some cherished concepts promoted here, and defended... rather strenuously.  And if someone posts an answer with a clear and simple explanation for my question, I'll dump mine.

The force is still a download on the horizontal stab, even with an aft center of gravity.
There's a reason that a CG envelope exists.
Stick trim force may be reduced to zero, but this does not imply loss of a tail down force.
In the case of the 172 previously mentioned, consider a simple wings-level unaccelerated straight-ahead power-off stall.  The pitch-down moment that occurs with the stall break, when the downwash off the wing is altered by airflow separation and the download on the horizontal stab decreases, is evidence of the download in the first place; this should occur at all points in the CG envelope (not just at the front of the envelope).
Consider also the consequence of a tailplane stall, typically associated with tailplane icing, or icing conditions, and the resulting loss of pitch authority.  Which way does the aircraft pitch, if loaded within the CG envelope?

You may be unfamiliar with the debate going on here for quite some time over the upforce downforce issue and it is quite nuanced.  I tend to be alone in your camp in here (my question is intended to address logical gaps I perceive in the other camp's arguments, which I need to see satisfied before I'll go along and agree that my perception is wrong).  There are some experienced designers here who know their stuff, are in the other camp, and I'm slowly ironing things out in my own mind.  Great fun and really interesting.

I'm not familiar with the turf war here, but am very familiar with the hangar-debate about upload and download, as well as the principles of stability common to most common general aviation and commercial transport category aircraft.  Stick force or trim force has everything to do with trimmed condition for a given speed and loading, but does not speak to upload or download on the horizontal stab.  Want a spanner in the works, consider the load on each surface of the Piaggio P.180.  That should cause some speculation.


text:
Are exoskeleton wings technically possible?
While there have been a robot simulating a bird, could be there also an exoskeleton wing system for a human?
Main design considerations:
a) only human powered
b) using an additional, or exclusive source of power
I consider b) is more trivial because it's near to more or less attaching a human to a big bionic bird similar to mentioned above.
So we concentrate here on a).

Human powered or not ?

@Antzi thanks for your question, I think only human powered is more challenging to explore, or? I have added this consideration to the question.

Flapping or not? Note that human powered flight has been attempted extensively and the most recent results point to a fit human being able to sustain slow flight in a highly optimized propeller-driven craft, and not for long

Yes, and it has been done.
In 2010, the first human-powered ornithopter recorded its first flight.
It took a very light aircraft and a lot of power from the pilot, about 650 W, so the flight only lasted 20 seconds. It only sustained level flight, requiring assistance to take off.
To put this into perspective, a good power target for cyclists is 250 W. A top competitor will put out 300 W over a race and 400 W over 10-15 minutes, and over 1000 W for 10-20 seconds. So it's possible (not certain) that a top Tour de France cyclist could hold it in the air for a couple minutes, or even have the stamina for a takeoff. But it's still more power than any human can sustain long enough to cover a mile.
Meanwhile, the record for propeller-driven human powered aircraft, without flapping wings, currently holds at 115 km. This was achieved with autonomous launch.
The ornithopter in question isn't exactly an "exoskeleton", though, because an exoskeleton is not optimal for the job. Human arms are unsuitable for flying or sustaining high power output in general, so all successful human-powered aircraft derive power from the pilot's legs. Typically it's a cycling motion, but a straight leg press machine motion is used in this particular case.

You may add a date as "9 year ago" won't be valid in few years while the rest of the answer still remain valid.

@Camille Goudeseune: Some good points in the edits, but the qualifiers in the post were important. E.g. I expect that a champion athlete could hold it for a couple minutes, but it's uncertain. I've incorporated those of your edit suggestions that I agree with.

Speaking about mechanically amplified wing that follows the movement of the human hand (something along the lines of hydraulic amplifiers that move control surfaces of large aircraft), most likely should not be any significant problem. Large birds beat wings somewhat 3 times per second (source), wings for the human may require even less. A human hand can beat up and down in such a frequency given exoskeleton provides the power.
It may be possible to imagine a machine where pilot waves hands freely in the cockpit with only sensors attached to them, and the moving wings are separate. This movement would be more complex than just up and down. Without tail surfaces, this would be more like a bat flies, and bats use they very long fingers to give the wings always the required shape. Probably the sensors should track the pilot fingers, palm position, adjusting the shape of the wing.
A bat can fly into cave and mount on the ceiling chassis up. Probably no helicopter (or some small model of it) could do.
I think this would be an amazing device to fly. Maybe someone could try to build at least a simulator - fly like a dragon!

For b), doubtful.  Not if exoskeleton has its conventional meaning, augmenting the wearer's gestures rather than performing gestures commanded by something like buttons on a joystick.
Uninterruptedly flapping anything bigger than your hands becomes tiring after just a few minutes, even with straps to counter the weight of your hands or arms or whatever.
Although the Kinect supported a few flap-your-arms games, that audience has completely returned to its keyboards and gamepads.  Button mashing corresponds to finger flapping.
Guitarists and drummers and pianists can play for an hour nonstop, but their motions are again minimal:  fingerwork; twisting the forearm (strumming); little elbow and shoulder motion.


text:
Effect of wing flexible in stall
How true is it that the deflection (due to large aspect ratio, flexible wing )  of a wing has a certain effect on the stall characteristics?  I read about this in the case of large wingspan gliders.  I was considering using an aerodynamic or geometric twist in the worst case but I read about this and wanted to see if for gliders this could be used to their advantage.
General Aviation Aircraft Design, by Snorri Gudmundsson:

Sailplane wingtips at a high AOA will flex substantially due to aeroelastic effects. This effectively unloads the wingtip and loads up the center portion of the wing, which then stalls before the wingtip. For this reason, sailplane usually have good stall characteristics.

Welcome to Aviation SE, Andy!  Let's narrow down what you're asking.  What does "deflection of a wing" mean?  Angle of attack, twist, camber due to ailerons or flaps?  Also, is "stall characteristics" the behavior of the airplane near stall, or just the wing's L/D curve?  (Please edit your question, instead of replying in another comment.)

Snorri's authoritative quote seems to answer your own question.  "Yes, a wing that twists spanwise changes the AoA along the span, thus changing which parts stall sooner, thus changing the airplane's stall characteristics."

Wing flex is an important consideration for most airplanes. Even the wing of a modern airliner will flex substantially, both in bending (tips up) and in torsion (tips twisted). Wing flex will also reduce aileron effectiveness at high speed; that is why Boeing used to add high-speed ailerons to their airliners.
Wing flex comes in two varieties: Bending means that the lift force will curve the wing tips up while the moment caused by the chordwise distance between the lift force and the elastic line of the wing will twist the wing. The high camber of glider airfoils causes large variations in the chordwise position of the center of lift, and since the effect of this torsion adds up over span, long aspect ratio wings are especially prone to torsion.
However, the largest torsion happens to occur at small angles of attack, going down as angle of attack is increased. Since the weight of the glider is constant, the same goes for the lift force, so torsion is inversely proportional to angle of attack. Stalling happens at high angle of attack when torsion is low, so it contributes only a little to the stall characteristics of gliders. However, in an accelerated stall or in a tight circle, when the wing has to produce more lift in proportion to the load factor, torsion can become significant and will unload the tips, reducing bending and increasing the stall margin on the outer wing.
Flaps have a larger effect on stall characteristics: For low speed flying they are lowered more on the inner wing, so it will contribute more to total lift and unload the tips. In the ASW-22 or the ASH-25 with flaps set to "L", the ailerons will actually move trailing edge up. If you fly those gliders at slightly more than approach speed, the wing tips will bend down from the negative lift on the outer wing.
See here for an answer demonstrating the effect of bending on stall characteristics.
Mr. Gudmundsson isn't directly wrong, but I think he exaggerates the effects of wing flex if he attributes the good stall characteristics of gliders to them. More important is careful airfoil selection and planform design: Glider designers know that their products will spend a lot of time near stall so they use airfoils with docile stall behavior and select taper and twist such that the tips will stall much later than the inner wing.

Why is it that "the largest torsion happens to occur at small angles of attack, going down as angle of attack is increased"? My intuition would be, more AoA = more twisting, no AoA, no twist. Where is my thinking wrong there?

Thank you.  I was looking at the possibility of applying an aerodynamic twist, however I could hardly find another airfoil that performed equal to the design lift coefficient and well it doesn't have flaps either, that's why I considered that possibility.  There is the other thing about vortex generators but I can't support it well, so I was looking for "clearer" ways.  The glider we are looking at has a 20 m wingspan and for the moment we plan to use a single profile.  Do you think it would be appropriate to use other methods to treat the loss effect?

@andyv: what do you have in mind exactly? Maybe a better answer can be given if you explain better what you want to create

Well.  It is a university project and it is to create a training sailplane however we have little time.  I already selected the profile that best suited the lift coefficient and offered a good maximum lift coefficient and everything that comes with the commitment between phases of the mission.

However, when we are working on the Schuemman platform (leading edge with taper and straight trailing edge) which is the classic one used in sailplane, due to the taper we have the detail of the stall at the tip where The aileron is going to be placed and that is why we were looking at how to do it (we had the option of using the straight leading edge and the trailing edge with the tapers, which does offer good stall characteristics) however the section where the aileron would go It could cost work due to the taper.

Well, at first we thought about using two profiles but it will take time to analyze and support the uses.  So our last option was to use a geometric twist of approximately 1.5° but we would adjust the spar then, hence the question of how feasible it is to use the wing's own deflection to treat the stall.  There are other devices but they entail a certain complexity that we cannot afford.

@RalphJ Your mistake is to assume the same dynamic pressure. I assume the same lift, so dynamic pressure has to compensate what is missing in AoA. Just as in real life.

@PeterKämpf Ah!  So, for a given lift, the higher AoA means less airspeed, and less torsion, than getting the same lift at a lower AoA and higher airspeed, yes?  That makes good sense, thanks!

@andyv: being an university project don't you have a professor following you? Or advising you where to find the needed information (i.e. not from random people on the internet)?


text:
Why does the nose pitch up during power reduction in a T-tail with rear mounted engines?
For the CRJ700 for example, reducing thrust causes a pitch up. Why is that?

It may be because the engines create a bit of nose down pitch at higher power based on the thrust vector being slightly higher than the CG.

The engines are mounted above the vertical center of gravity position. That means there is a vertical moment arm from the thrust which causes a pitch down when thrust is increased and a pitch up when thrust is decreased.
From a stabilized trimmed condition, e.g. during straight and level flight your pitch trim counteracts the nose down effect. When you then suddenly reduce thrust that trimmed condition is no longer in balance and the stabilizer pitches the plane up due to the reduced nose down moment.
Most airliners with engines mounted under the wing have the exact opposite effect. There the engines are below the center of gravity and any increase in thrust causes a pitch up and a decrease causes a pitch down.
There are other effects at play as well such as the jet wash and airspeed trim stability but the vertical moment arm is the largest factor during a sudden change in thrust.

At least in the CRJ 700, I'd suggest that it's not primarily from the engines being mounted above the CG. They are probably a little above the CG, but only minutely.
But when we look at a side view, we quickly see that the engine's orientation almost certainly have more effect:

The engines are well behind the CG, and the front of each engine is tilted up by what looks like around 10 degrees. That being the case, I'd expect increased thrust to give a nose-down pitch, regardless of the height at which they were mounted.
Photo Credit
https://www.delta.com/us/en/aircraft/bombardier/crj-700

The engine orientation is there to align them with the local flow and should be around 4 degrees. Both vertical position and angle add up to create the pitch effect.


text:
What is the speed of the approach lights rabbit?
If you were flying past the rabbit and passed each bulb as it lit, what speed would you be flying at? The rabbit is 2400 feet long, spaced at 100 foot increments and consists of 15 lights. The math doesn't add up on that but it's here on the FAA Visual Guidance Lighting Systems. It flashes twice per second but it's not as simple as 2400 feet in half a second, because the lights don't immediately start over when it reaches the end. So the simulated speed is a bit faster than 3272 mph (2400 ft / 0.5 sec).
https://www.faa.gov/about/office_org/headquarters_offices/ato/service_units/techops/navservices/lsg/als

Why do you mention "mph" in your question? In aviation "miles" are always nautical miles (except for WX) and speed is in knots. Anyway, I can try to remember to ask my Tech Ops guy next time I see him—I'll get back to you.

The rabbit length is only 1400 feet for the ALSF-2; 2400 feet is the total system distance. For SSALR and MALSR the rabbit is 800 feet long. For MALSF it is only 400 feet long.

Since the rabbit is made of light, the obvious answer is "the speed of ...".   ;)

@randomhead - re "and speed is in knots" - not always- I'm looking at an ASI calibrated in mph right now; expect to see the resulting raw data featured in a future answer.  And the aircraft is quite capable of outrunning the rabbit.  (Well, the actual warm-blooded kind.)

The diagram in the reference distinguishes between runway threshold and landing threshold, FWIW...

There are several different systems of Sequenced Flashing Lights, which is the official name for the "rabbit." Generally these flashing lights are spaced 200 feet apart.
FAA Order JO 6850.2B is the current specification for "Visual Guidance Lighting Systems." It describes the following systems which include flashing lights:

ALSF-2. Fifteen flashing lights spaced at 100-foot intervals from THR–1000' to THR–2400' (total: 1400').
SSALR. Five flashing lights spaced at 200-foot intervals from THR–1600' to THR-2400' (total: 800').
MALSR. Five flashing lights spaced at 200-foot intervals from THR–1600' to THR-2400' (total: 800').
MALSF. Three flashing lights spaced at 200-foot intervals from THR–1000' to THR–1400' (total: 400').
ODALS. Five flashing lights spaced at 300-foot intervals from THR–300' to THR–1500' (total: 1200').

All of the above systems are described as "flash[ing] in sequence toward the threshold at the rate of twice per second" except for ODALS, which is not described as flashing toward the threshold at all; however from watching a YouTube video depicting an ODALS installation and timing with a stopwatch, I judge that ODALS flashes at a "rate" of once per second. (You can compare this to a video of an ALSF-2 and see the difference.) But this tells us only how often the sequence begins; it does not tell us how long the sequence takes (although there is an upper bound of 0.5 seconds, as there is only one sequence active at any one time).
Your question asks: What is the speed at which the "flash" travels down the length of the SFLs? We can answer this definitely if we know two pieces of information: the distance between successive flashers, and the time interval between successive flashes. We know the first but we do not know the second.
As @abelenky describes, we can assume that the sequence takes the entire 0.5 seconds to elapse, in other words, that it takes the rabbit the entire 0.5 seconds to travel 1400 feet:
$$ \frac{1400 \mathrm{~ft}}{0.5 \mathrm{~sec}} \cdot \frac{1 \mathrm{~NM}}{6076 \mathrm{~ft}} \cdot \frac{3600 \mathrm{~sec}}{1 \mathrm{~hr}} = 1650  \mathrm{~kt} $$
This gives us a correct lower bound on the rabbit's speed for the ALSF-2. If we performed this calculation on the MASLR or MALSF we would reach a lower lower bound, because we would be assuming that the rabbit takes the entire 0.5 seconds to travel 800 feet or 400 feet respectively. But my strong suspicion is that the rabbit travels at the same speed no matter the type of system, and the only difference is the "pause" at the end of each sequence—a shorter system will complete its sequence in less time and require more of a post-sequence delay until the next half-second.
The definitive answer can only come from FAA documentation describing the light-to-light timing, and I have not been able to find that easily. I will ask around and see what I can do. In the meantime, I have taken the ALSF-2 YouTube video from above and stepped through it frame by frame; I can confirm that it takes twelve frames (0.5 seconds) for the sequence to re-start, and further that it takes only seven frames for the sequence to elapse. Thus I can estimate the rabbit's speed:
$$ \frac{1400 \mathrm{~ft}}{7/24 \mathrm{~sec}} \cdot \frac{1 \mathrm{~NM}}{6076 \mathrm{~ft}} \cdot \frac{3600 \mathrm{~sec}}{1 \mathrm{~hr}} = 2840  \mathrm{~kt} $$
If we think about in terms of what values are likely to be set by a human, 1400 feet in 7/24 of a second comes out to 4800 feet per second, or in other words (close to) 20 milliseconds per 100 feet—which equals 2960 kt.

What’s really fascinating is that is Mach 4.4!!  I’ve never tried to visualize the speed of sound like that.  To get Mach 1 that .5 sec would need to increase to about 1.25sec!

The "flashers" part of the system is only 1400 feet long (the 15 red triangles in the diagram on the page you linked to).
If the lights blink twice per second, we can set a lower bound on the speed.
I'm assuming only one light is on at a time, and the first light in the sequence flashes as soon as the last light goes off.
Outline

The first light flashes at time 0.00
The last light flashes at some time no later than 0.5, 1,400 feet away
The first light flashes again at exactly time 0.50 (making twice per second)Starting the the second "rabbit run".

So, the "Rabbit Run" covers 1,400 feet in 1/2 second or less.
(It could be faster, if there is some gap in time between the first and second rabbit-run in a one second interval.)
So, I think the math is:
$$ \frac{1400 \mathrm{~ft}}{0.5 \mathrm{~sec}} \cdot \frac{1 \mathrm{~NM}}{6076 \mathrm{~ft}} \cdot \frac{3600 \mathrm{~sec}}{1 \mathrm{~hr}} = 1658  \mathrm{~kt} $$
To get a more precise estimate, I would want to know the time delay between the last light turning off, and the first light turning on again.

It would also still meet the spec of twice per second if the first light flashed at time 0.00 and again at time 0.99.
That is not how I originally interpreted it, but it seems valid.
In that case, the math would lead to about: $829 \mathrm{~kt}$

Yep, that looks to be about right.  Looking at the FAA description, and doing the math a bit differently, the sweep speed of the ALS approach system lighting is 2845 kt.

@ThomasPerry unfortunately that is not quite right... I am in the middle of composing an answer explaining why. It may be close, however. Note that the correct distance for the SFLs of an ALSF-2 is 1400ft, not 2400ft.

@ThomasPerry I must eat my words a little. By stepping through a real-life video take of an ALSF-2 I come up with a sequence time of seven frames, which gives a speed of 2844kt (prior to sig-fig-ization).